Leetcode:Word Ladder

url:

https://leetcode.com/problems/word-ladder/#/description

题目大意：

Given two words (beginWord and endWord), and a dictionary’s word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

Only one letter can be changed at a time.Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

For example,

Given:
beginWord = “hit”
endWord = “cog”
wordList = [“hot”,”dot”,”dog”,”lot”,”log”,”cog”]

As one shortest transformation is “hit” -> “hot” -> “dot” -> “dog” -> “cog”,
return its length 5.

Note:

Return 0 if there is no such transformation sequence.All words have the same length.All words contain only lowercase alphabetic characters.You may assume no duplicates in the word list.You may assume beginWord and endWord are non-empty and are not the same.

解题思路：

以beginWord为起点，用’a’-‘z’替换当前值中每一个位置的值，以此进行BFS。具体代码如下：

static final char[] chars = new char[]{'a','b','c','d','e','f','g',            'h','i','j','k','l','m','n',            'o','p','q','r','s','t','u','v','w','x','y','z'};    public int ladderLength(String beginWord, String endWord, List<String> wordList) {        Set<String> worldSet = new HashSet<String>(wordList);//将列list转为set是为了后续进行contain和remove更快，如果list会超时        Queue<String> queue = new LinkedList<String>();        queue.add(beginWord);        if(worldSet.contains(beginWord)){            worldSet.remove(beginWord);        }        if(!worldSet.contains(endWord))//如果endWord不在wordlist中则直接返回0            return 0;        int level = 0;//记录路径长度        while(!queue.isEmpty()){            int size = queue.size();            for(int i=0;i<size;i++){                String cur = queue.remove();                if(endWord.equals(cur)){                    return level+1;                }                char[] curChars = cur.toCharArray();                char oldChar;                String temp=null;                for(int k=0;k<curChars.length;k++){                    oldChar = curChars[k];                    for(int j=0;j<26;j++){                        curChars[k] = chars[j];//每一个位置进行字母替换                        temp = new String(curChars);                        if(!cur.equals(temp)&&worldSet.contains(temp)){                            queue.add(temp);                            worldSet.remove(temp);                        }                    }                    curChars[k] = oldChar;//将这个位置的字母还原                }            }            level++;        }        return 0;    }

运行结果：